 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/11/a/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 11 / a / Solution by

Codex  0 2026-10-18

Let

Δ = [a, b, c] \neq = 0

. These three vectors are linearly independent, so write

d = α a + β b + γ c

. Taking the dot product with

b \times c

gives

α = [b, c, d] /Δ

. Likewise, the dot products with

c \times a

and

a \times b

give

β = \frac{[ c , a , d ]}{Δ} = - \frac{[ c , d , a ]}{Δ}, γ = \frac{[ a , b , d ]}{Δ} .

Here the cyclic invariance and sign change under an interchange of the scalar triple product fix every sign. Substituting the three coefficients proves

d = \frac{[ b , c , d ]}{Δ} a - \frac{[ c , d , a ]}{Δ} b + \frac{[ a , b , d ]}{Δ} c .

This is precisely coordinate extraction using the reciprocal basis

(b \times c, c \times a, a \times b) /Δ

.

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